∫_-1^1log≤ft(x+√(x^2+1))dx is equal to

The correct answer is Option (3) → 0

Since the limits are symmetric, it helps to see whether \( f(x)=\log\left(x+\sqrt{x^2+1}\right) \) is odd or even.

    \( \displaystyle f(-x)=\log\left(-x+\sqrt{x^2+1}\right) \)

    \( \displaystyle =\log\left(\frac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x}\right) \)

    \( \displaystyle =\log\left(\frac{(x^2+1)-x^2}{x+\sqrt{x^2+1}}\right) \)

    \( \displaystyle =\log\left(\frac{1}{x+\sqrt{x^2+1}}\right) \)

    \( \displaystyle =-f(x) \)

So f is an odd function, and for an odd function \( \int_{-a}^{a}f(x)\,dx=0 \).

    \( \displaystyle \int_{-1}^{1}\displaystyle\log\left(x+\sqrt{x^2+1}\right)dx=0 \)

Tip: \( \log\left(x+\sqrt{x^2+1}\right) \) is a standard odd function worth remembering: its integral over any interval [−a, a] is 0.